`int e^(-3x)sin5x dx` Find the indefinite integral

Expert Answers

An illustration of the letter 'A' in a speech bubbles

We shall use partial integration:

`int u dv=uv-int v du`  

Therefore, we have

`int e^(-3x)sin5xdx=|[u=e^(-3x),dv=sin5xdx],[du=-3e^(-3x)dx,v=-1/5cos5x]|=`

`-1/5e^(-3x)cos5x-3/5int e^(-3x)cos5xdx=|[u=e^(-3x),dv=cos5xdx],[du=-3e^(-3x)dx,v=1/5sin5x]|=` 

`-1/5e^(-3x)cos5x-3/25e^(-3x)sin5x-9/25inte^(-3x)sin5xdx`                                                                                  

We can see that we have the same integral as the one we've started with. In other words we have the following equation

`int e^(-3x)sin5xdx=-1/5e^(-3x)cos5x-3/25e^(-3x)sin5x`

-`9/25inte^(-3x)sin5xdx`

Let us add `9/25int e^(-3x)sin5xdx` to the whole equation.

`34/25int e^(-3x)sin5xdx=-1/5e^(-3x)cos5x-3/25e^(-3x)sin5x`  

Now we only need to multiply the whole equation by `25/34` to obtain the solution to our starting problem.

`int e^(-3x)sin5xdx=-5/34e^(-3x)cos5x-3/34e^(-3x)sin5x+c,` `c in RR`                                                                             

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial